Description: Obsolete version of dfss2 as of 16-May-2024. (Contributed by NM, 8-Jan-2002) (Proof modification is discouraged.) (New usage is discouraged.)
Ref | Expression | ||
---|---|---|---|
Assertion | dfss2OLD | ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfss | ⊢ ( 𝐴 ⊆ 𝐵 ↔ 𝐴 = ( 𝐴 ∩ 𝐵 ) ) | |
2 | df-in | ⊢ ( 𝐴 ∩ 𝐵 ) = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } | |
3 | 2 | eqeq2i | ⊢ ( 𝐴 = ( 𝐴 ∩ 𝐵 ) ↔ 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } ) |
4 | abeq2 | ⊢ ( 𝐴 = { 𝑥 ∣ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) } ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ) ) | |
5 | 1 3 4 | 3bitri | ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ) ) |
6 | pm4.71 | ⊢ ( ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ) ) | |
7 | 6 | albii | ⊢ ( ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ 𝑥 ∈ 𝐵 ) ) ) |
8 | 5 7 | bitr4i | ⊢ ( 𝐴 ⊆ 𝐵 ↔ ∀ 𝑥 ( 𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵 ) ) |